Question

question 10 of 20 what are the coordinates of the foci of the conic section shown below? $\frac{(y + 2)^2}{16}-\frac{(x - 3)^2}{9}=1$ a. $(3,-2pm5)$ b. $(-2,3pm5)$ c. $(-2pm5,3)$ d. $(-2pmsqrt{7},3)$

Explanation:

Step1: Identify the form of hyperbola

The given equation $\frac{(y + 2)^2}{16}-\frac{(x - 3)^2}{9}=1$ is of the form $\frac{(y - k)^2}{a^2}-\frac{(x - h)^2}{b^2}=1$, where $(h,k)$ is the center of the hyperbola. Here, $h = 3$, $k=-2$, $a^2 = 16$ (so $a = 4$), and $b^2=9$ (so $b = 3$).

Step2: Calculate $c$

For a hyperbola, the relationship between $a$, $b$, and $c$ is $c^2=a^2 + b^2$. Substitute $a = 4$ and $b = 3$ into the formula: $c^2=16 + 9=25$, so $c = 5$.

Step3: Find the foci

For a hyperbola of the form $\frac{(y - k)^2}{a^2}-\frac{(x - h)^2}{b^2}=1$, the foci are located at $(h,k\pm c)$. Substitute $h = 3$, $k=-2$, and $c = 5$ into the formula. The foci are $(3,-2\pm5)$.

Answer:

A. $(3, - 2\pm5)$